Given that \( A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix} \), we aim to determine matrix \( A \).
The inverse of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated as:
\[
A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
Applying this formula to the given \( A^{-1} \):
\[
A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & 1 \\ -3 & 2 \end{bmatrix}
\]
The determinant of \( A^{-1} \) is computed as:
\[
\det(A^{-1}) = \left(\frac{1}{7}\right)^2 (2 \cdot 2 - 1 \cdot (-3)) = \frac{1}{49} (4 + 3) = \frac{7}{49} = \frac{1}{7}
\]
The inverse of \( A^{-1} \), which is \( A \), is then found using:
\[
A = \frac{1}{\det(A^{-1})} \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} = 7 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}
\]
Consequently, the solution is:
\[
\boxed{A = 7 \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix}}
\]